problem: Linear equation question the question is:
$$(y^2+xy^3)\mathrm dx + (5y^2-xy+y^3\sin(y))\mathrm dy = 0$$
can any body tell me how to solve this linear equation?? when I tried to solve this the expression of integrating factor becomes too much difficult, may be i calculated it wrong... 
Any help will be appreciated. 
Thanks! 
 A: First I'd divide through by $y^2$ to make your life easier. Then the partial differential equation for the integrating factor becomes
$$xu+\frac{\partial u}{\partial y}(1+xy)=-\frac{1}{y}u+\frac{\partial u}{\partial x}(t-\frac{x}{y}+y\sin y)\;.$$
That happens to have a solution with $\frac{\partial u}{\partial x}=0$, so you can determine an integrating factor $u(y)$ from it.
A: $(y^2+xy^3)~dx+(5y^2-xy+y^3\sin y)~dy=0$
$(xy^3+y^2)~dx=(xy-5y^2-y^3\sin y)~dy$
$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=\dfrac{x}{y^2}-\dfrac{5}{y}-\sin y$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=\dfrac{1}{y^2}\left(u-\dfrac{1}{y}\right)-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}+\dfrac{u}{y^2}=\dfrac{u}{y^2}-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u\dfrac{du}{dy}=-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$
$u~du=\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\int u~du=\int\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$
$\dfrac{u^2}{2}=\dfrac{1}{2y^2}-5\ln y+\cos y+c$
$\left(x+\dfrac{1}{y}\right)^2=\dfrac{1}{y^2}-10\ln y+2\cos y+C$
A: \begin{cases} M(x,y)=y^2+xy^3 \\\\
N(x,y)=5y^2-xy+y^3\sin y\\\\
\end{cases}
\begin{cases} \frac{\partial M}{\partial y}=2y+3xy^2 \\\\
\frac{\partial N}{\partial x}=-y\\\\
\end{cases}
so non-exact
$p(y)=\frac{3y(1+xy)}{-y^{2}(1+xy)}$
$p(y)=\frac{-3}{y}$
$μ(y)=e^{\int\frac{-3}{y}dy}=e^{\ln(y^{-3})}=y^{-3}$  , integrating factor
$(y^{-1}+x)~dx+(5y^{-1}-xy^{-2}+\sin y)~dy=0$
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=-y^{-2}$
so exact
$f(x,y)=\int(y^{-1}+x)~dx+h(y)=\frac{x^{2}}{2}+\frac{x}{y}+h(y)$
$-xy^{-2}+h'(y)=5y^{-1}-xy^{-2}+\sin y$
$h(y)=\int (5y^{-1}+\sin y)~dy=5\ln(y)-\cos y$
$$f(x,y)=\frac{x^{2}}{2}+\frac{x}{y}+5\ln(y)-\cos y=c$$
